phyisics
posted by Carolina .
an orangutan throws a coconut vertically upward at the foot of a clif 40 m high while his mate simulaneously drops another coconut from the top of the clif. the two coconuts collide at an altitude of 20 m what is the initial velocity of the coconut that was thrown upward?
i know the equations
x= position = .5a t^2 + v (velocity) initial t + x initial (pos)
v= at+v initial
a=a
please help thanks!!

phyisics 
bobpursley
Go to the bottom coconut
20=Vi*t1/2 g t^2
Now go to the top coconut
20=401/2 g t^2
solve for t in the second equation, put it in the first, solve for vi 
phyisics 
Carolina
waht is g?

phyisics 
Carolina
and what do you mean by top and bottom?
thanks 
phyisics 
bobpursley
top of clift, bottom of cliff
g is 9.8m/s^2, the acceleration of gravity. 
phyisics 
Carolina
why didyou use
these equations
Go to the bottom coconut
20=Vi*t1/2 g t^2
Now go to the top coconut
20=401/2 g t^2 
phyisics 
bobpursley
You were given distance, those are the distance equation.
finalposition=initial position+vi*t+ 1/2 a*t^2
memorize that. g= is negative(downward), so make the last term  1/2 9.8 t^2, some just write  4.9t^2 
phyisics 
Heisenberg
If objects acted simultaneously their relationship is time. The time is obtain because the time of collision between the coconut is the same.
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